Let y(x) be a solution of the differential equation $(1+{{e}^{x}})y'+y{{e}^{x}}=1$ . If y(0) = 2, then which of the following statements is (are) true?
$\begin{align}
  & \text{a) }\text{y(-4) = 0} \\
 & b)\text{ y(-2) = 0} \\
 & \text{c) y(x) has a critical point in the interval (-1,0)} \\
 & \text{d) y(x) has no critical point in the interval (-1,0)} \\
\end{align}$

Hint: Now we are given with the equation that can be written in the form of an ordinary linear differential equation $\dfrac{dy}{dx}+P(x)y=Q(x)$ . Hence we will find an integrating factor for this which is given by ${{e}^{\int{P(x)dx}}}$ . Now multiplying the equation with the integrating factor we will get an equation which can be easily solved as LHS will be in the form $\dfrac{d(u.v)}{dx}$ . Now we will integrate the equation and get the required general Solution of Differential equation. Further we will use the condition y(0) = 2 to find the integrating Constant and hence we will now have the particular solution of the differential equation.

Complete step by step answer:
Now we are given with the equation $(1+{{e}^{x}})y'+y{{e}^{x}}=1$ .
Now let us divide the equation by $(1+{{e}^{x}})$
$y'+\dfrac{{{e}^{x}}}{(1+{{e}^{x}})}y=\dfrac{1}{(1+{{e}^{x}})}$
Now the equation is in the form of an ordinary linear differential equation.
Now we can see that in the given equations we cannot separate the variables hence, we will use the integrating factor method to solve the equation.
Now according to this method integrating factor of the equation $\dfrac{dy}{dx}+P(x)y=Q(x)$ is given by ${{e}^{\int{P(x)dx}}}$
Hence in the given equation the Integrating factor is ${{e}^{\int{\dfrac{{{e}^{x}}}{(1+{{e}^{x}})}dx}}}.....................(1)$
Now first let us first solve the integral part. Consider $\int{\dfrac{{{e}^{x}}}{(1+{{e}^{x}})}dx}$
Now let us take $(1+{{e}^{x}})=t\Rightarrow {{e}^{x}}dx=dt$
Hence we get
\[\begin{align}
  & \int{\dfrac{{{e}^{x}}}{(1+{{e}^{x}})}dx}=\int{\dfrac{dt}{t}} \\
 & =\ln t+C \\
\end{align}\]
Now let us again resubstitute the value of t in the equation
$\int{\dfrac{{{e}^{x}}}{(1+{{e}^{x}})}dx}=\ln (1+{{e}^{x}})$
Now substituting this in equation (1) we get integrating factor equal to
${{e}^{\int{P(x)dx}}}={{e}^{\int{\dfrac{{{e}^{x}}}{(1+{{e}^{x}})}dx}}}={{e}^{\ln (1+{{e}^{x}})}}=(1+{{e}^{x}})$
Now multiplying the integral factor $(1+{{e}^{x}})$to the given equation $y'+\dfrac{{{e}^{x}}}{(1+{{e}^{x}})}y=\dfrac{1}{(1+{{e}^{x}})}$ we get
$y'(1+{{e}^{x}})+{{e}^{x}}y=1$
Now note that LHS is actually nothing but $\dfrac{d(y.(1+{{e}^{x}}))}{dx}=1$ as we know $\dfrac{d(u.v)}{dx}=u'v+v'u$
Now let us integrate the equation $\dfrac{d(y.(1+{{e}^{x}}))}{dx}=1$ on both side.
$\begin{align}
  & \int{d(y.(1+{{e}^{x}}))=\int{1}dx} \\
 & \Rightarrow y(1+{{e}^{x}})=x+C \\
\end{align}$
Hence now we have the solution of the differential equation as $y(1+{{e}^{x}})=x+C..............(2)$ .
Now we know that y (0) = 2. We will use this condition to find the value of C.
Let us substitute x = 0 in the equation. Hence we get
$\begin{align}
  & 2(1+{{e}^{0}})=0+C \\
 & \Rightarrow 2(1+1)=C \\
 & \Rightarrow C=4 \\
\end{align}$
Hence, now we have the value of C is 4. Substituting this in equation (2) we get
$\begin{align}
  & y(1+{{e}^{x}})=x+4 \\
 & \Rightarrow y=\dfrac{(x+4)}{(1+{{e}^{x}})} \\
\end{align}$
Hence now we have the function
$y=\dfrac{(x+4)}{(1+{{e}^{x}})}...............(3)$
Now we will have a critical point if $(1+{{e}^{x}})=0$ but for this ${{e}^{x}}=-1$ which is not possible as ${{e}^{x}}>0$

Hence the y(x) has no critical point in $(-1,0)$
Now let us find the solution for y(x) = 0 hence substituting y = 0 in equation (3)
$\begin{align}
  & 0=\dfrac{(x+4)}{(1+{{e}^{x}})} \\
 & x+4=0 \\
 & x=-4 \\
\end{align}$
Hence we have y(-4) = 0

So, the correct answer is “Option A and D”.

Note: While taking the integrating factor we compare the equation with $\dfrac{dy}{dx}+P(x)y=Q(x)$ to find P(x). here note that the coefficient of $\dfrac{dy}{dx}$ is 1. Hence if there is a function multiplied to $\dfrac{dy}{dx}$
We will divide the whole equation with the function to make it look like general form which is
$\dfrac{dy}{dx}+P(x)y=Q(x)$